What is the Maximum Energy and Velocity in Spring Oscillation?

[Naldo6]

A 0.70 kg mass on a spring oscilates horizontally with a little friction acording to the
x= 0.0500 cos (2.10 t) where x is in meter and t is in seconds. Find the maximum energy stored in the spring during an oscilation. Find the maximum velocity of the mass.

V_max = w*A= (0.05)*(2.10)= 0.105 m/s

But the maximum energy E=(1/2)(m)(v_max ^2)= (1/2)(0.70)(0.105)^2= 3.86 x 10^-3 J

Is this problem correct?... or i use the wrong equations...

 

[Doc Al]

Looks OK to me.

 

[thomate1]

Your calculations appear to be correct. The maximum energy stored in the spring during an oscillation can be calculated using the equation E = (1/2)kA^2, where k is the spring constant and A is the amplitude of the oscillation. In this case, the amplitude is equal to 0.05 m and the spring constant can be calculated using the equation k = mω^2, where m is the mass and ω is the angular frequency (in this case, ω = 2.10 rad/s). Plugging in these values, we get E = (1/2)(0.70)(2.10)^2(0.05)^2 = 3.86 x 10^-3 J, which is the same as your result.

As for the maximum velocity of the mass, you correctly used the equation v_max = ωA. This represents the maximum speed at which the mass is moving during the oscillation. In this case, the maximum velocity is equal to 0.105 m/s.

Overall, your approach and calculations seem to be correct. However, it is always important to double-check your equations and units to ensure accuracy. Additionally, it is good practice to include units in your calculations to avoid any confusion.